What is the integral from 0 to 1 of x(sqrt(x)-x) ?

The integral from 0 to 1 of x(sqrt(x)-x) is 1/15

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integral from 0 to 1 of x(sqrt(x)-x)

\int_{0}^{1} x (x - x) d x

\frac{1}{15}

Decimal Notation

0.06666 \dots

Solution steps

\int_{0}^{1} x (x - x) d x

Expand

x (x - x) : x^{\frac{3}{2}} - x^{2}

= \int_{0}^{1} x^{\frac{3}{2}} - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} x^{\frac{3}{2}} d x - \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{\frac{3}{2}} d x = \frac{2}{5}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= \frac{2}{5} - \frac{1}{3}

Simplify

\frac{2}{5} - \frac{1}{3} : \frac{1}{15}

= \frac{1}{15}

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What is the integral from 0 to 1 of x(sqrt(x)-x) ?
The integral from 0 to 1 of x(sqrt(x)-x) is 1/15